What is the sum from n=4 to infinity of 1/(n^2-4n+3) ?

The sum from n=4 to infinity of 1/(n^2-4n+3) is converges

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sum from n=4 to infinity of 1/(n^2-4n+3)

n = 4 \sum \infty \frac{1}{n ^{2} - 4 n + 3}

converges

Solution steps

n = 4 \sum \infty \frac{1}{n ^{2} - 4 n + 3}

Apply Series Comparison Test:

converges

= converges

n = 2 \sum \infty \frac{1}{n + 1} - \frac{1}{n}

n = 1 \sum \infty \frac{n + 1}{2}

n = 1 \sum \infty \frac{5}{( ( - 4 ) ^{n} )}

k = 0 \sum \infty (\frac{1}{10})^{k}

k = 1 \sum \infty 3 (- \frac{1}{5})^{3 k}

What is the sum from n=4 to infinity of 1/(n^2-4n+3) ?
The sum from n=4 to infinity of 1/(n^2-4n+3) is converges